The perfect gas law relates the pressure p ($$\frac{N}{m^2}$$) , absolute temperature T (K), mass n (moles) and volume V ($$m^3$$) of a gas. It states that
$$ pV = nRT $$
where the constant R is the gas constant. The value of R for air is 286.7 Nm/kgK. Suppose air is contained in a chamber at room temperature ($$ 0^oC $$ = 273K).
A. Create a function that computes the gas pressure p in $$\frac{N}{m^2}$$ from input arguments V and n. Ensure your function accepts V as a range of values.

Guruprasad R.

Answer:

The function in MATLAB to solve this question is presented below with comments to help with understanding.
%Task (Function to output values of P)
function [P] = task3_evaluatePressure (V,m) %P is the output variable, and (V,m) are the input %variables
R = 286.7; %Given
T = 293; %Given
P = m*R*T./V; %./ is the element wise division element as V is a range of values {follows from formula}
end

Linear Algebra

TutorMe

Question:

$$ W= {(x,y) \in R^2 | xy \geq 0 } $$
Which of the following statements is true:
a) W is a subspace
b) W is not a subspace as it isn't closed under vector addition
c) W is not a subspace as it isn't closed under scalar multiplication
d) W is not a subspace as it doesn't contain the zero vector

Guruprasad R.

Answer:

To determine if W is a subspace or not, there are 3 major criteria one must verify:
i) W should have the zero vector
$$ \underline{Proof}: $$
Let element (0,0) belong to W. Does it abide by the conditions imposed? Here, the condition is (0,0) $$ \in R^2 $$ and $$ 0*0 \geq 0 $$.
In this case, both the conditions are satisfied, which implies it does contain the zero vector.
ii) W should be closed under vector addition
$$ \underline{Proof}: $$
Let {$$ (x_1, y_1), (x_2, y_2) $$} belong to the W space. This implies that both the conditions imposed must be met by the elements {$$ (x_1, y_1), (x_2, y_2) $$}.
In other words: $$ (x_1, y_1) \in R^2; (x_2, y_2) \in R^2 $$ and $$ x_1*y_1 \geq 0; x_2*y_2 \geq 0$$
Now, to verify the vector addition of the 2 elements lie within the W-space.
Vector addition of the 2 elements ==> $$ (x_1 + x_2, y_1 + y_2) $$
To show they belong to the W-space, we need to prove that $$ (x_1+x_2)*(y_1+y_2) \geq 0 $$
$$
\begin{equation}
= (x_1 + x_2)*(y_1 + y_2) \\
= (x_1*y_1 + x_2*y_2 + x_1*y_2 + x_2*y_1) \\
\end{equation} $$
We know that the first 2 terms $$ (x_1*y_1 + x_2*y_2 ) \geq 0 $$, but the last 2 terms could be less than 0 - resulting in the expression evaluating to a negative value.
As there are cases when the vector addition isn't closed, W is not a subspace.
iii) W should be closed under scalar multiplication
$$ \underline{Proof}: $$
Let {$$ (x_1, y_1) $$} belong to the W space. This implies that both the conditions imposed must be met by the element {$$ (x_1, y_1) $$}.
In other words: $$ (x_1, y_1) \in R^2 $$ and $$ x_1*y_1 \geq 0 $$
Now, to verify if the scalar multiplication of the element lies within the W-space.
Let 'c' be a scalar and it's product with the element is: $$ (c*x_1, c*y_1) $$
To show this element belongs to the W -space, we need to show that:
$$
\begin{equation}
(c*x_1) * (c*y_1) \geq 0 \\
= c*c * x_1*y_1 \\
= c^2 *(x_1 * y_1)
\end{equation}
$$
For any scalar $$ c \in R $$, $$ c^2 \geq 0 $$ AND $$ x_1*y_1 \geq 0 $$. Therefore the product $$ c^2 *(x_1 * y_1) \geq 0 $$
This shows that W is closed under scalar multiplication.
Given this analysis: we can see that the only true statement is ;
(b) - W is not a subspace as it isn't closed under vector addition

Before, we jump to evaluating the expression by applying relevant formulas, the limits of the integral should be noticed.
The integral of an expression (here - $$\frac{(x^2-e^x)}{sin^2(x)} $$) implies the evaluation of the area under the curve between the limits of integration. As in the question, if the limits are from 5 to 5, whatever be the function (curve), the area under the curve will be 0.
Therefore, without having to evaluate the complex expression, we arrive at the answer - 0.

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